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I am looking for an elementary derivation of the formula for the area of a geodesic triangle lying in a surface of constant curvature $\kappa$, depending on the angles and side length.

Of course, the formula can easily be derived from the Gauss–Bonnet formula to be $$A = \frac{1}{\kappa}(\alpha + \beta + \gamma - \pi)$$ for $\kappa \neq 0$. However, I would like to have an elementary geometric proof.

Does anybody know a reference?

Ricardo Andrade
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Matthias Ludewig
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    For $\kappa > 0$, I believe you can use inclusion-exclusion with 3 geodesics on a sphere. – S. Carnahan May 12 '13 at 12:39
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    For $\kappa<0$ Gauss had a similar inclusion-exclusion proof: It is in his collected works (letter to Wolfgang/Farkas Bolyai). – Misha May 12 '13 at 21:31
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    For both cases, there are visual proofs, which can be considered elementary: http://mathoverflow.net/questions/8846/proofs-without-words/16064#16064 and http://mathoverflow.net/questions/8846/proofs-without-words/24774#24774 – Igor Khavkine May 13 '13 at 00:30
  • There are some proofs by considering bigons (for spherical geometry) and subdivisions of an ideal triangle (in hyperbolic geometry). You may also consult J. Ratcliffe's book "Foundations of hyperbolic manifolds", some of the first chapters. However, the area of a bigon or ideal triangle is computed by "not very elementary" means there. I agree with Anton, all "elementary proofs" are cheating up to a certain extent. – SashaKolpakov Aug 31 '13 at 23:25

2 Answers2

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M. Berger, Geometrie, vol. V. MR0536874

Edit. Let me sketch a proof for the spherical triangle. Let the sphere have area $4\pi$. First you derive the area of digon. It is $2\alpha$, where $\alpha$ is the angle, by completely elementary reasons. Now consider a triangle. Extend its sides to three full great circles. These three circles make several digons and two equal triangles (the second one is centrally symmetric to the original one). Make a picture showing how these three circles partition the sphere. As the areas of all digons are known the area of a triangle is simply derived by the exclusion-inclusion formula!

Notice: this proof is truly elementary in the sense that it only uses the existence of the area for a digon and triangle, its invariance with respect to rotations, and finite additivity. Euclid COULD give a rigorous proof of this. As rigorous as his investigation of areas of Euclidean triangles.

Alexandre Eremenko
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All the "elementary derivations of that type" are cheating (it may look nice but it proves nothing).

The only elementary way to introduce area is adding it as an axiom (which is already kind of cheating). You have to say that there is a additive area-functional on the set of all polygons. Then you probably want to prove that this functional is unique (or include it in the same axiom).

It remains to notice that your functional $A$ satisfies the same properties and nothing left to prove.

Anton Petrunin
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    You would need some normalization axiom, in order to distinguish between proportional measures. Do you have a specific one in mind? – Sergei Ivanov May 12 '13 at 22:38
  • @Sergei, yes sure, all I wanted to say is that if one knows what is area and curvature then there is nothing to prove. – Anton Petrunin May 13 '13 at 02:17
  • I don't understand what you mean. I have a Riemannian metric of constant curvature on a Surface. This gives me curvature and a volume density, so I know both what curvature and area is and I can in theory calculate what the area of a geodesic triangle is. Where do I need an axiom? – Matthias Ludewig May 13 '13 at 07:12
  • @Kofi, you ask for an elementary derivation. For me "Riemannian metric" and "integral" are not elementary and the geometry as it was used to be covered in the school (but not any more) is elementary. – Anton Petrunin May 13 '13 at 17:02
  • P.S. surface of constant curvature κ are spheres plane or Lobachevsky plane. All these things are "elementary" for me. – Anton Petrunin May 13 '13 at 17:45
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    Anton: I disagree with what you say. The area of a TRIANGLE is an elementary notion. (The theory of areas of triangles in Euclid is completely rigorous, by all modern standards.) And the formula has a really elementary proof. – Alexandre Eremenko May 14 '13 at 20:47
  • @Alexandre, I do not see what exactly you disagree with. A rigorous intro to area from the axioms takes 20-40 pages, and once it is done the formula is already proved. So these sort of "proofs" confuse poorly educated students and they prove nothing to those who know what area is. – Anton Petrunin May 15 '13 at 04:13
  • What about the scissors congruence definition af area of simple polygons ? – BS. May 16 '13 at 09:46
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    @BS, check this question http://mathoverflow.net/questions/119953/definition-of-area I would be very happy if you know a better answer. – Anton Petrunin May 16 '13 at 18:33
  • Anton, sorry I looked at 119953 and I don't understand your objection. In elementary geometry we deal with areas of polygons. The area is defined by a) finite additivity, b) monotonicity, invariance with respect to motion, c) the area of the unit square is 1. From this it is easy to derive that the area of a polygon exists and is unique. And I believe Euclid did it rigorously. Kiselev (who wrote the common Russian high school geometry text) did it rigorously, and I studied this in 8-th grade. What's wrong with all this? – Alexandre Eremenko May 16 '13 at 20:41
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    @Alexandre, Euclid (and Kiselev) did not prove the existence, essentially they add the existence as an axiom, but they did not say that it is an "axiom". This axiom follows from the rest of axioms, but it takes 20 pages at least. Instead of unit square you have to use other normalization (which essentially defines curvature). A rigourous way to introduce area given in "Elementary Geometry From An Advanced Standpoint" by Moise 35 pages Euclidean plane onlyy,and in "Geometry: A Metric Approach with Models" by Millman and Parker 40 pages neutral plane and contains a gap. – Anton Petrunin May 16 '13 at 23:37